\(n_{Fe}=\frac{16,8}{56}=0,3\left(mol\right)\)
\(Fe_2O_3+3CO\underrightarrow{^{to}}2Fe+3CO_2\)
\(n_{CO2}=\frac{3}{2}n_{Fe}=0,45\left(mol\right)\)
\(\Rightarrow m_{CO2}=0,45.44=19,8\left(g\right)\)
Fe2O3+3CO---->2Fe+3CO2
n Fe=16,8/56=0,3(mol)
n CO2=3/2n Fe=0,45(g)
m CO2=0,45.44=19,8(g)\
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